SOLUTION: 5n^2+19n+12 2v^2+11v+5 2n^2+5n+2 m^+2m-24 x^2-4x+24 k^2-13k+40

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 5n^2+19n+12 2v^2+11v+5 2n^2+5n+2 m^+2m-24 x^2-4x+24 k^2-13k+40      Log On


   

Question 420164: 5n^2+19n+12
2v^2+11v+5
2n^2+5n+2
m^+2m-24
x^2-4x+24
k^2-13k+40
Answer by dnanos(83) About Me  (Show Source):
You can put this solution on YOUR website!
%285n%5E2%2B19n%2B12%29= multiply 5*12=60...Search among the factors of 60 to find two of them having sum 19 ( that is...4*15=60 4+15=19)
after that we have:
%285n%5E2%2B15n%2B4n%2B12%29=
%285n%28n%2B3%29%2B4%28n%2B3%29%29=
%28%28n%2B3%29%285n%2B4%29%29
etc.....
Same way:
2v%5E2%2B11v%2B5+= We have 2*5=10 1*10=10 1+10=11
2v%5E2%2Bv%2B10v%2B5=
v%282v%2B1%29%2B5%282v%2B1%29=
%282v%2B1%29%28v%2B5%29
For a better way we use the sqrt%28b%5E2-4ac%29 of the general formula ax%5E2%2Bbx%2Bc and if we find b%5E2-4ac%290 then we cannot factorize.
I.e.the above:
x%5E2-4x%2B24
a=1,b=-4,c=24
and b%5E2-4%2A1%2A24=%28-4%29%5E2-4%2A24=-80%3C0